Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


Difficult problem. Been thinking about it for a few hours now. Pretty sure it's beyond my ability. Very frustrating to show that the limit even exists.

Help, please. Either I'm not smart enough to solve this, or I haven't learned enough to solve this. And I want to know which!

share|cite|improve this question
just playing around, $\lim_{x \to 1^-}\left( \frac{\lim_{n \to \infty }\left( 1-x^{2n}\right) - (1-x^{2})\log_2 \left( 1 \over 1-x\right)}{1-x^{2}}\right)$, this seem to go like $\lim_{n \to \infty} n - \log_2 n$ – Santosh Linkha Feb 6 '13 at 6:29

Write $x:=e^{-2^{\delta}}$. Then the desired limit is $\lim_{\delta\to-\infty} F(\delta)+\log_2 (1-e^{-2^{\delta}})$, where $$ F(\delta):=\sum_{n\ge 0} e^{-2^{\delta+n}}.$$ But if $$ G(\delta):=\sum_{n\ge 0} e^{-2^{\delta+n}}+\sum_{n<0} (e^{-2^{\delta+n}}-1) $$ then shifting the index of summation shows that $G(\delta+1)=G(\delta)-1$, so $G(\delta)+\delta$ has period $1$. Calling this periodic function $H(\delta)$, then, \begin{eqnarray*} F(\delta)+\log_2 (1-e^{-2^{\delta}}) &=& H(\delta) -\delta + \log_2 (1-e^{-2^{\delta}}) - \sum_{n<0} (e^{-2^{\delta+n}}-1)\\ &=&H(\delta)+O(2^\delta),\qquad\delta\to-\infty. \end{eqnarray*} Computing the periodic function $H$ numerically shows that it is not a constant. Therefore, the function whose limit is being taken is oscillatory, so the limit does not exist.

share|cite|improve this answer

This is NOT a solution, but I think that others can benefit from my failed attempt. Recall that $\log_2 a=\frac{\log a}{\log 2}$, and that $\log(1-x)=-\sum_{n=1}^\infty\frac{x^n}n$ for $-1\leq x<1$, so your limit becomes


The series above can be rewritten as $\frac1{\log2}\sum_{k=1}^\infty a_kx^k$, where

$$a_k=\begin{cases} -\frac1k,\ &\style{font-family:inherit;}{\text{if}}\ k\ \style{font-family:inherit;}{\text{is not a power of}}\ 2;\\\log2-\frac1k,\ &\style{font-family:inherit;}{\text{if}}\ k=2^m.\end{cases}$$

We can try to use Abel's theorem, so we consider $\sum_{k=1}^\infty a_k$. Luckily, if this series converges, say to $L$, then the desired limit is equal to $1+\frac L{\log2}\,$. Given $r\geq1$, then we have $2^m\leq r<2^{m+1}$, with $m\geq1$. Then the $r$-th partial sum of this series is equal to


where $H_r$ stands for the $r$-th harmonic number. It is well-known that

$$\lim_{r\to\infty}H_r-\log r=\gamma\quad\style{font-family:inherit;}{\text{(Euler-Mascheroni constant)}}\,,$$

so $$\sum_{k=1}^ra_k=\log(2^m)-\log r-(H_r-\log r)=\log\Bigl(\frac{2^m}r\Bigr)-(H_r-\log r\bigr)\,.$$

Now the bad news: the second term clearly tends to $-\gamma$ when $r\to\infty$, but unfortunately the first term oscillates between $\log 1=0$ (when $r=2^m$) and $\bigl(\log\frac12\bigr)^+$ (when $r=2^{m+1}-1$), so $\sum_{k=1}^\infty a_k$ diverges.

share|cite|improve this answer
Perhaps $\sum a_k$ is Cesàro summable? Abel's theorem would still apply then. – Antonio Vargas Feb 6 '13 at 7:18
@AntonioVargas Thanks for the suggestion, though from David Moews' answer we can conclude that $\sum a_k$ is not Cesàro summable, either. By the way, where can I find a proof of this more general version of Abel's theorem? Almighty Wikipedia does not make any reference to this generalization. – Matemáticos Chibchas Feb 10 '13 at 1:04
I learned the result from Hardy's book Divergent Series. Unfortunately I don't have a copy handy so I can't point you to a particular page. – Antonio Vargas Feb 10 '13 at 2:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.